The twisted product is a group iff the factor set is a two-cocycle

Assume $f$ is a normalized two-cocycle. Using the normalization equations from [F1], one checks directly from [F2] that $(0,1)$ is a two-sided identity. The inverse of $(m,g)$ is $(-g^{-1}\cdot m-g^{-1}\cdot f(g,g^{-1}),g^{-1})\mathrm{.}$

Still under the cocycle hypothesis, compute both products $((m,g)(n,h))(r,k)$ and $(m,g)((n,h)(r,k))$ from [F2]. Their second coordinates are both $ghk$, and equality of the first coordinates is exactly the cocycle equation from [F1]. So the law is associative.

Conversely, suppose the twisted law makes $M\times G$ a group with identity $(0,1)$. Comparing $(m,g)(0,1)$ and $(0,1)(m,g)$ with $(m,g)$ forces $f(g,1)=f(1,g)=0$. Comparing $((0,g)(0,h))(0,k)$ and $(0,g)((0,h)(0,k))$ then yields the cocycle equation. Hence $f$ is a normalized two-cocycle.

Step 1.2 proves the forward implication, and step 1.3 proves the reverse implication.